Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Wave optics

  1. Apr 7, 2004 #1
    I know how to do these, I think I just somehow fudged up the calculations:

    1)A pair of narrow slits is illuminated with red light of wavelength [itex]\lambda[/itex] = 633 nm. The slits are separated by 0.10 mm center-to-center. (a) What is the angular separation of the interference maxima near the center of the pattern? (b) How far apart are neighboring maxima if they are observed on a wall 6.55 m away from the slits?

    2)Calculate the width of the central maximum in the single-slit diffractiion pattern of yello light of [itex]\lambda[/itex] = 589.0 nm by a slit 0.250 mm wide viewed on a screen 2.00 m away.

    Just to give u an idea of how bad I think I screwed up, I got an answer of around 50 m for the second problem.
  2. jcsd
  3. Apr 7, 2004 #2


    User Avatar
    Science Advisor
    Gold Member

    #1.a.) For the double-slit, we know that maxima occur at integer values of m for:
    [tex] d\sin\theta = m\lambda [/tex]
    (d is slit separation)
    at small values of θ (using a distant screen approximation), the angular separation of consecutive maxima is given by solving for θ when m = 1:
    [tex] \theta = \frac{\lambda}{d} [/tex]
    b.) to find the distance these are separated on the wall, use θ~y/D:
    [tex] y = \frac{\lambda D}d [/tex]
    where D is the screen distance.

    #2. If you're looking for the full width of the central maximum, can't you find that by looking at the position of the two minima that are on either side of the central maximum? The formula for a maximum is given by
    [tex] a\sin\theta = m\lambda [/tex]
    where a is the aperture width. m is an integer for a max, but is an odd multiple of 1/2 for a min, so the width of the central max should be defined by m = +1/2 and m = -1/2
    Because of the symmetry, this is the same as twice the angle about m = 1/2:
    [tex]\sin\theta = \frac{\lambda}{2a} [/tex]
    Use the small angle approximation (if you want to) to find:
    [tex] \sin\theta \sim \theta \sim \tan\theta = \frac y D = \frac{\lambda}{2a} [/tex]
    So the width, w, should be:
    [tex] w = 2y = 2D\frac{\lambda}{2a} = \frac{\lambda D}a [/tex]
    When you plug in the numbers, I get a little less than half a cm.

    These questions are usually pretty straightforward. The most likely source of error is a mistake with the units.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook